if the base has length x,
area = 1/2 * x * x√3 = √3/2 x^2 = 50/√3
x = 10
so, y = 10√3
so, the line is
x/10 + y/(10√3) = 1
find the equation of a straight line on which the perpendicular from the origin makes an angle of 30 degree with the x-axis and which forms a triangle of are 50/root3 with the axes
2 answers
oops. x = 10/√3
y = 10
x/(10√3) + y/10 = 1
y = 10
x/(10√3) + y/10 = 1